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\title{Five Regime Problem}

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\maketitle{}

In this attempt, I discard the regime that both fossil fuel and
renewable energy are used. So here we have a transition point
instead of a period. as long as the shadow
prices of the two energy are equal, fossil fuel use drops to zero and
renewable energy use becomes a positive number.

\section{Differential Equation System}

In the next following sections, I will specify the differential
equations for each regime in a chronological order.
\subsection{Regime 1: Fossil Fuel Only, $R>0$, $B=0$, $n>0$, $j=0$}
At the beginning, only fossil fuel is used because of its lower
cost. We need solve five DEs for $\{k, S, N, \sigma, \lambda\}$:

\begin{equation}
\dot{k}=i-\delta k
\label{eq:Reg1_keq}
\end{equation}
\begin{equation}
\dot{S}=QAk
\label{eq:Reg1_Seq}
\end{equation}
\begin{equation}
\dot{N}=n
\label{eq:Reg1_Neq}
\end{equation}
\begin{equation}
\dot{\sigma}=\beta \sigma +\lambda \frac{\partial g}{\partial S}Ak
\label{eq:Reg1_sigeq}
\end{equation}
\begin{equation}
\dot{\lambda}=\lambda(\beta+\delta+(g(S,N)-1)A)-\sigma QA
\label{eq:Reg1_lameq}
\end{equation}
together with the exogenous population growth $Q=Q_{0}e^{\pi t}$.

The two control variables $i$ and $n$ can be solved using the
following two equations:

\begin{equation}
\begin{split}
& n\lambda\left( \frac{\partial ^{2}g}{\partial N^{2}}k-2\frac{\partial g}{%
\partial N}\right) =\\&\lambda \left[ \frac{\partial g}{\partial N}[k(\delta
+g(S,N)A-A+\frac{\sigma QA}{\lambda })+\lambda ^{-1/\gamma }]-\frac{\partial
^{2}g}{\partial S\partial N}QAk^{2}\right] -\sigma \pi Q
\end{split}
\label{eq:n_soln}
\end{equation}%

\begin{equation}
i=Ak[1-g(S,N)]-\lambda ^{-1/\gamma }-n  \label{eq:i_fossregime}
\end{equation}%



\subsection{Regime 2: Renewable Only, $R=0$, $B>0$, $n=0$, $j=0$ }

Now consider the beginning of the renewable energy regime where
j=0. Note that this regime may not exist. We should remember to
incorporate this uncertainty in the computer program.

In this case, we need solve four equations for $\{k, H, \lambda, \eta\}$:

\begin{gather}
\dot{k}=Ak[1-(\Gamma_1+H)^{-\alpha}]-\lambda^{-1/\gamma}-\delta k  \label{eq:Reg2a_keq} \\
\dot{H}=Ak  \label{eq:Reg2a_Heq} \\
\dot{\lambda}=\lambda \left[ \beta +\delta -A(1-(\Gamma _{1}+H)^{-\alpha })\right]-\eta A  \label{eq:Reg2a_lameq}\\
\dot{\eta}=\beta\eta-\lambda\alpha(\Gamma _{1}+H)^{-\alpha-1}Ak \label{eq:Reg2a_etaeq}
\end{gather}

\subsection{Regime 3: Renewable Only, $R=0$, $B>0$, $n=0$, $j>0$ }

In this regime, We start to invest in R\&D in renewable sector, $j>0$.
Three DEs for $\{k, H, \lambda\}$ need to be solved here:

\begin{gather}
\dot{k}=i-\delta k  \label{eq:Reg2_keq} \\
\dot{H}=Ak(1+\psi j)  \label{eq:Reg2_Heq}\\
\dot{\lambda}=\lambda \left[ \beta +\delta -A(1-(\Gamma _{1}+H)^{-\alpha })-\frac{1+\psi j}{\psi k}\right]  \label{eq:Reg2_lameq}
\end{gather}

The two equations below can then be
solved for $i$ and $j$ given current values for $k,H$ and $\lambda$:

\begin{equation}
i+j=Ak(1-(\Gamma _{1}+H)^{-\alpha })-\lambda ^{-1/\gamma }
\label{eq:budgetreg2}
\end{equation}

\begin{equation}
\begin{split}
& \left[ 2[\delta -A(1-(\Gamma _{1}+H)^{-\alpha })]+3\alpha\psi (Ak)^2(\Gamma_{1}+H)^{-\alpha -1}\right] i+ \\
& \left[ \frac{\lambda ^{-1/\gamma }}{\gamma k}-\alpha\psi (Ak)^2(\Gamma _{1}+H)^{-\alpha -1}[2+(1+\alpha)Ak^2(\Gamma _{1}+H)^{-1}]\right] j \\
& =\left[ 2[\delta -A(1-(\Gamma _{1}+H)^{-\alpha })]+3\alpha\psi (Ak)^2(\Gamma_{1}+H)^{-\alpha -1}\right] \delta k \\
&+\alpha (Ak)^2(\Gamma _{1}+H)^{-\alpha -1}[2+(1+\alpha)Ak^2(\Gamma _{1}+H)^{-1}]\\
& +\frac{\lambda ^{-1/\gamma }}{\gamma }[\beta +\delta -A(1-(\Gamma_{1}+H)^{-\alpha })-\frac{1}{\psi k}]
\end{split}
\label{eq:ijrelreg2}
\end{equation}


\subsection{Regime 4: Renewable Only, $R=0$, $B>0$, $n=0$, $j=0$}

The DE system in this regime is the same as that in Regime 2. Also
like Regime 2, this regime may not exist either. Remember we should
allow this in the program.

\subsection{Regime 5: Renewable Only, $R=0$, $B=0$, $n>0$, $j=0$,
  $p=\Gamma_2$}

We have two DEs in this regime for $\{k,\lambda\}$, which can be
solved analytically.

The solution can be written as:

\begin{equation}
\lambda =\bar{K}e^{\bar{A}t}  \label{eq:term_lam_sol}
\end{equation}%
for some constant $\bar{K}$ yet to be determined, and

\begin{equation}
k = \frac{\gamma \bar{K}^{-1/\gamma}e^{-\bar{A}t/\gamma}}{\beta\gamma-\bar{A}%
(\gamma-1)}  \label{eq:Term_k_sol}
\end{equation}
with $\lambda$ given by \eqref{eq:term_lam_sol} and $\bar{K}$ is a constant
yet to be determined.


\section{Boundary Conditions}

As long as we clarify all the DEs we are going to use to solve the
model, it's time to determine the boundary conditons and stop events. The discussion
is as below, for both backward and forward problem.

\subsection{To Solve Backward}

Denote the transition time for each regime are $T_1$, $T_2$, $T_3$ and
$T_4$. The terminal condition for Regime 4 at $T_4$ is $\{k_{T_4},
H_{T_4}, \lambda_{T_4}, \eta_{T_4}\}$, in which $k_{T_4}$ and $\lambda_{T_4}$ can be
solved in Regime 5, $H_{T_4}=\Gamma_2^{-1/\alpha}-\Gamma_1$, and
$eta_{T_4}=0$. The stop event for regime 4 is $\eta=\lambda/\psi*A*k)$.

If Regime 4 exists, the terminal condition for Regime 3 at $T_3$ is $\{k_{T_3},
H_{T_3}, \lambda_{T_3}\}$, which is solved in Regime 4. If not, we have
$k_{T_3}$ and $\lambda_{T_3}$ solved in Regime 5, and
$H_{T_3}=\Gamma_2^{-1/\alpha}-\Gamma_1$.

The stop event for regime 3 is $j=0$. If this event doesn't happen,
use $H=0$ instead.

The terminal condition for Regime 2 at $T_2$ is $\{k_{T_2},
H_{T_2}, \lambda_{T_2}, \eta_{T_2}\}$, in which the first three are solved
in Regime 3, and $\eta_{T_2}$ is given by $\eta=\lambda/\psi*A*k)$. The
stop event for regime 2 is $H=0$.

At $T_1$, we need the terminal condition $\{k_{T_1}, S_{T_1}, N_{T_1},
\sigma_{T_1} \lambda_{T_1}\}$, in which $k_{T_1}$ and $\lambda_{T_1}$
can be solved in Regime 2 or 3 (depending on that Regime 3 exists or
not), $\sigma_{T_1}=0$, $N_{T_1}$ is the initial guess and $S_{T_1}$ is
given by the following equation:

\begin{equation}
\left[ \delta +g(S,N)A-\frac{\partial g}{\partial N}R-A\right] \lambda
=\sigma QA  \label{eq:qdoteqnudot}
\end{equation}%

To solve the model, we have three guesses: $T_4$, the unknown constant in
Regime 5 $\bar{K}$, and $N_{T_1}$. Three targets need to be hit are: $S_0 = 0$, $N_0 = 0$, and $k_0 = 3.6071282734$.





\subsection{To Solve Forward}



\end{document}
